This is far from a complete answer, but too long for a comment. Specifically, I'll adress the beginning of 2', and the question of the relation between $A(\mathbb N)$ and $A(\mathbb Z)$ that was raised in the comments.

In both cases I'm assuming that your definition of $A(M)$ is about finite $M$-sets, and not arbitrary $M$-sets (whatever that would mean).

2' : The $\pi_0$ of a group completion is always the (ordinary) group completion of $\pi_0$ by some abstract nonsense with adjoints, so $\pi_0 (|C_\mathbb N|^{grp}) \cong (\pi_0|C_\mathbb N|)^{grp}$, and $\pi_0|C_\mathbb N|$ is exactly the semi-ring that you're using to define $A(\mathbb N)$, so they do coincide.

For the relation between $A(\mathbb Z)$ and $A(\mathbb N)$, note that there is an obvious "inclusion" map $A(\mathbb Z)\to A(\mathbb N)$. In fact, it is really an inclusion because it has a retraction $A(\mathbb N)\to A(\mathbb Z)$, which sends a finite set with endomorphism $(X,f)$ to the essential image $im_\infty(f)$ with the induced endomorphism $f_\infty$. This essential image is $\bigcap_n f^n(X)$ which is clearly stable under $f$, and on which $f$ is surjective, so (by finiteness) bijective.

The construction $(X,f)\mapsto (im_\infty(f),f_\infty)$ is functorial, in particular it is so when restricting to isomorphisms. Furthermore, it preserves disjoint unions and cartesian products so it induces a map of semi-($E_\infty$-)ring $E_\infty$ spaces $|C_\mathbb N|\to |C_\mathbb Z|$ and so a retraction of rings $A(\mathbb N)\to A(\mathbb Z)$ (but also of commutative ring spectra before passing to $\pi_0$).

However, $A(\mathbb Z)\to A(\mathbb N)$ is not an isomorphism. Indeed, consider the following group map $A(\mathbb N)\to \mathbb Z$, defined by sending $(X,f)$ to the cardinal of $X\setminus im_\infty(f)$. Because $im_\infty(f)$ preserves disjoint unions, this sends disjoint unions to sums and it sends an automorphism $f$ to $0$, however it is not $0$ on all $A(\mathbb N)$.

(if you take "THH" of finite sets instead of $K$-theory of finite sets with endomorphisms, $(im_\infty(f),f_\infty)$ is all you can recover, cf. my answer here, but this crucially uses cyclic invariance, so an identification of $(X,f\circ g)$ with $(Y, g\circ f)$)

More generally, you have all the maps sending $X$ to the cardinal of $f^n(X)$ which are equal to the augmentation map on $A(\mathbb Z)$, but not on the whole of $A(\mathbb N)$. I would guess that the collection of these maps induces an injection $A(\mathbb N)\to A(\mathbb Z)\times \prod_\mathbb N\mathbb Z$  , but that is just a guess.

$\DeclareMathOperator\im{im}$This is far from a complete answer, but too long for a comment. Specifically, I'll adress the beginning of 2', and the question of the relation between $A(\mathbb N)$ and $A(\mathbb Z)$ that was raised in the comments.

In both cases I'm assuming that your definition of $A(M)$ is about finite $M$-sets, and not arbitrary $M$-sets (whatever that would mean).

2' : The $\pi_0$ of a group completion is always the (ordinary) group completion of $\pi_0$ by some abstract nonsense with adjoints, so $\pi_0 (\lvert C_\mathbb N\rvert^\text{grp}) \cong (\pi_0\lvert C_\mathbb N\rvert)^\text{grp}$, and $\pi_0\lvert C_\mathbb N\rvert$ is exactly the semi-ring that you're using to define $A(\mathbb N)$, so they do coincide.

For the relation between $A(\mathbb Z)$ and $A(\mathbb N)$, note that there is an obvious "inclusion" map $A(\mathbb Z)\to A(\mathbb N)$. In fact, it is really an inclusion because it has a retraction $A(\mathbb N)\to A(\mathbb Z)$, which sends a finite set with endomorphism $(X,f)$ to the essential image $\im_\infty(f)$ with the induced endomorphism $f_\infty$. This essential image is $\bigcap_n f^n(X)$ which is clearly stable under $f$, and on which $f$ is surjective, so (by finiteness) bijective.

The construction $(X,f)\mapsto (\im_\infty(f),f_\infty)$ is functorial, in particular it is so when restricting to isomorphisms. Furthermore, it preserves disjoint unions and cartesian products so it induces a map of semi-($E_\infty$-)ring $E_\infty$ spaces $\lvert C_\mathbb N\rvert\to \lvert C_\mathbb Z\rvert$ and so a retraction of rings $A(\mathbb N)\to A(\mathbb Z)$ (but also of commutative ring spectra before passing to $\pi_0$).

However, $A(\mathbb Z)\to A(\mathbb N)$ is not an isomorphism. Indeed, consider the following group map $A(\mathbb N)\to \mathbb Z$, defined by sending $(X,f)$ to the cardinal of $X\setminus \im_\infty(f)$. Because $\im_\infty(f)$ preserves disjoint unions, this sends disjoint unions to sums and it sends an automorphism $f$ to $0$, however it is not $0$ on all $A(\mathbb N)$.

(If you take "THH" of finite sets instead of $K$-theory of finite sets with endomorphisms, $(\im_\infty(f),f_\infty)$ is all you can recover, cf. my answer to What are the conjugacy classes of the category of ($\kappa$-small) sets?, but this crucially uses cyclic invariance, so an identification of $(X,f\circ g)$ with $(Y, g\circ f)$.)

More generally, you have all the maps sending $X$ to the cardinal of $f^n(X)$ which are equal to the augmentation map on $A(\mathbb Z)$, but not on the whole of $A(\mathbb N)$. I would guess that the collection of these maps induces an injection $A(\mathbb N)\to A(\mathbb Z)\times \prod_\mathbb N\mathbb Z$, but that is just a guess.